Answer
Total area of the rose is $\frac{\pi }{2}$.
Since the circumscribed circle has unit radius, its area is $\pi$. Therefore, the total area of the rose is equal to one-half the area of the circumscribed circle.
Work Step by Step
First, we find the limits of integration for one leaf.
When $\theta=0$, we have $r=0$, so the rose starts at the origin. When $\theta = \frac{\pi }{4}$, we have $r=1$. From Figure 17, we see that the curve touches the circumscribed circle at $\theta = \frac{\pi }{4}$. Thus, the circle has radius $1$. When $\theta = \frac{\pi }{2}$, we have $r=0$. So, the curve traces out one leaf for the interval $0 \le \theta \le \frac{\pi }{2}$.
Using Eq. (2) of Theorem 1, the area of one leaf is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\sin ^2}2\theta {\rm{d}}\theta $
Since ${\sin ^2}2\theta = \frac{1}{2}\left( {1 - \cos 4\theta } \right)$, the integral becomes
$area = \frac{1}{4}\cdot\mathop \smallint \limits_0^{\pi /2} \left( {1 - \cos 4\theta } \right){\rm{d}}\theta $
$area = \frac{1}{4}\left( {\theta - \frac{1}{4}\sin 4\theta } \right)|_0^{\pi /2} = \frac{1}{4}\left( {\frac{\pi }{2}} \right) = \frac{\pi }{8}$
By symmetry, the total area of the rose is four times the area of one leaf. So, total area is $\frac{\pi }{2}$.
Since the circumscribed circle has unit radius, its area is $\pi$. Therefore, the total area of the rose is equal to one-half the area of the circumscribed circle.
